Simplify and expand the following expression: $ \dfrac{3}{z - 7}+ \dfrac{1}{4z - 40}+ \dfrac{2}{z^2 - 17z + 70} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4z - 40} = \dfrac{1}{4(z - 10)}$ We can factor the quadratic in the third term: $ \dfrac{2}{z^2 - 17z + 70} = \dfrac{2}{(z - 7)(z - 10)}$ Now we have: $ \dfrac{3}{z - 7}+ \dfrac{1}{4(z - 10)}+ \dfrac{2}{(z - 7)(z - 10)} $ The least common multiple of the denominators is: $ (z - 7)(z - 10)$ In order to get the first term over $(z - 7)(z - 10)$ , multiply by $\dfrac{4(z - 10)}{4(z - 10)}$ $ \dfrac{3}{z - 7} \times \dfrac{4(z - 10)}{4(z - 10)} = \dfrac{12(z - 10)}{(z - 7)(z - 10)} $ In order to get the second term over $(z - 7)(z - 10)$ , multiply by $\dfrac{z - 7}{z - 7}$ $ \dfrac{1}{4(z - 10)} \times \dfrac{z - 7}{z - 7} = \dfrac{z - 7}{(z - 7)(z - 10)} $ In order to get the third term over $(z - 7)(z - 10)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{2}{(z - 7)(z - 10)} \times \dfrac{4}{4} = \dfrac{8}{(z - 7)(z - 10)} $ Now we have: $ \dfrac{12(z - 10)}{(z - 7)(z - 10)} + \dfrac{z - 7}{(z - 7)(z - 10)} + \dfrac{8}{(z - 7)(z - 10)} $ $ = \dfrac{ 12(z - 10) + z - 7 + 8} {(z - 7)(z - 10)} $ Expand: $ = \dfrac{12z - 120 + z - 7 + 8}{4z^2 - 68z + 280} $ $ = \dfrac{13z - 119}{4z^2 - 68z + 280}$